\[\theta ={{\tan }^{-1}}(2ta{{n}^{2}}\theta )-ta{{n}^{-1}}\left( \frac{1}{3}\tan \theta \right)\] then \[\tan \theta =\]

\[\theta ={{\tan }^{-1}}(2ta{{n}^{2}}\theta )-ta{{n}^{-1}}\left( \frac

1.	\[\theta ={{\tan }^{-1}}(2ta{{n}^{2}}\theta )-ta{{n}^{-1}}\left( \frac{1}{3}\tan \theta \right)\] then \[\tan \theta =\]
A.	\[-2\]
B.	\[-1\]
C.	\[2/3\]
D.	\[2\]
Answer» B. \[-1\]

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\[\theta ={{\tan }^{-1}}(2ta{{n}^{2}}\theta )-ta{{n}^{-1}}\left( \frac{1}{3}\tan \theta \right)\] then \[\tan \theta =\]

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